
\prob{0058}{从高到底边}

已知$\triangle ABC$的三顶点$A, B, C$所对三边长分别为$a, b, c$，三边所对应的高分别为$h_a, h_b, h_c$。若$h_a = 3\sqrt{15}, h_b = 4\sqrt{15}, h_c = 6\sqrt{15}$，求$a, b, c$。
\problabels{yellow/平面几何, green/长度问题}

\ans{$a = 32, b = 24, c = 16$}

\subsection{设比例} \label{subsec:0058-rt}

显然
\[ ah_a = bh_b = ch_c \Rightarrow 3a = 4b = 6c \]
设$a = 4k, b = 3k, c = 2k$，求出$k$即可。

\begin{figure}[htbp]
  \centering \image{0058-rt}
  \caption{方法~\ref{subsec:0058-rt} 图} \label{fig:0058-rt}
\end{figure}

如图~\ref{fig:0058-rt}，作$CH \perp AB$于$H$，显然有
\[ AH^2 + CH^2 = AC^2, BH^2 + CH^2 = BC^2 \]
设$AH = x, BH = y$，则
\[ \left\{ \begin{aligned}
  y - x &= AB = 2k \\
  x^2 + h_c^2 &= (3k)^2 = 9k^2 \\
  y^2 + h_c^2 &= (4k)^2 = 16k^2
\end{aligned} \right. \]
后两式相减得
\[ (y + x)(y - x) = 7k^2 \Rightarrow y + x = \frac72k \]
于是
\[ \left\{ \begin{aligned}
  y - x &= 2k \\
  y + x &= \frac72k
\end{aligned} \right.
\Rightarrow \left\{ \begin{aligned}
  x &= \frac34k \\
  y &= \frac{11}4k
\end{aligned} \right. \]

在$\rttri AHC$中，
\[ AH^2 + CH^2 = AC^2 \Rightarrow x^2 + h_c^2 = 9k^2 \]
即
\[ 9k^2 + 8640 = 144k^2 \Rightarrow k = 8 \]
于是
\[ a = 4k = 32, b = 3k = 24, c = 2k = 16 \]
